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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>We know that <span class="process-math">\(C_1 y_1\)</span> satisfies the ODE. We replace <span class="process-math">\(C_1\)</span> by a function of <span class="process-math">\(x\text{,}\)</span> say <span class="process-math">\(v(x)\text{.}\)</span> Then, we try to determine <span class="process-math">\(v(x)\)</span> to make <span class="process-math">\(y_2=v(x) y_1\)</span> be a solution which is also independent of <span class="process-math">\(y_1\text{.}\)</span> Substituting <span class="process-math">\(y_2=v(x) y_1\)</span> into the ODE, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;\quad ~ v^{\prime \prime} y_1+2 v^{\prime} y_1^{\prime}+v y_1^{\prime \prime}+b (v^{\prime} y_1+v y_1^{\prime})+c v y_1=0\\
&amp;\to~v (y_1^{\prime \prime}+b y_1^{\prime}+ c y_1)+v^{\prime \prime} y_1+2 v^{\prime} y_1^{\prime}+b v^{\prime}y_1=0
\\
&amp;\to~ 0+v^{\prime \prime} y_1+v^{\prime} (2 y_1^{\prime}+b y_1)=0\\
&amp;\to~v^{\prime \prime} e^{-\frac{b}{2}x}+v^{\prime} \cdot 0=0\\
&amp;\to~v^{\prime \prime}=0\\
&amp;\to~v(x)=d_1 x+d_2,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(d_1, d_2\)</span> are arbitrary constants.</p>
<span class="incontext"><a href="sec3_5.html#p-103" class="internal">in-context</a></span>
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